I have a file, the content of file has a string like this:
'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'
I want to check the file has this string. But when I use grep to check, It always return false. I try some ways:
grep "'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'" foo.txt
grep "'/ad/e','#'\.base64_decode\("ZXZhbA\=\="\)\.'\(\$zad\)', 'add'" foo.txt
str="'/ad/e','#'\.base64_decode\("ZXZhbA\=\="\)\.'\(\$zad\)', 'add'"
grep "$str" foo.txt
Can you help me? Maybe, another command line.
This is my case:
while read str; do
if [ ! -z "$str" ]; then
if grep -Fxq "$str" "$file_path"; then
do somthing
fi
fi
done < <(cat /usr/local/caotoc/db.dat)
Thank you so much!
First, you need to make sure the string is quoted properly. This is a bit of an art form, since your string contains both single and double quotes.
One thought would be to use read and a here-document to avoid having to escape anything.
Second, you need to use -F to perform exact string matching instead of more general regular-expression matching.
IFS= read -r str <<'EOF'
'/ad/e','#'.base64_decode("ZXZhbA==").'($zad)', 'add'
EOF
grep -F "$str" foo.txt
Based on the update, you can use a simple loop to read them one at a time.
while IFS= read -r str; do
grep -F "$str" foo.txt
done < /usr/local/caotoc/db.dat
You may be able to simply use the -f option to grep, which will cause grep to output lines from foo.txt that match any line from db.dat.
grep -f /usr/local/caotoc/db.dat -F foo.txt
Instead of trying to workaround regexes, the simplest way is to turn off regular expressions using -F (or --fixed-strings) option, which makes grep act like a simple string search
-F, --fixed-strings PATTERN is a set of newline-separated strings
like this:
grep -F "'/ad/e','#'.base64_decode(\"ZXZhbA==\").'(\$zad)', 'add'" test
Note: because of the shell, you still need to escape:
double quotes
dollar sign or else $zad is evaluated as an environment variable
Related
I am using the below code for replacing a string
inside a shell script.
echo $LINE | sed -e 's/12345678/"$replace"/g'
but it's getting replaced with $replace instead of the value of that variable.
Could anybody tell what went wrong?
If you want to interpret $replace, you should not use single quotes since they prevent variable substitution.
Try:
echo $LINE | sed -e "s/12345678/${replace}/g"
Transcript:
pax> export replace=987654321
pax> echo X123456789X | sed "s/123456789/${replace}/"
X987654321X
pax> _
Just be careful to ensure that ${replace} doesn't have any characters of significance to sed (like / for instance) since it will cause confusion unless escaped. But if, as you say, you're replacing one number with another, that shouldn't be a problem.
you can use the shell (bash/ksh).
$ var="12345678abc"
$ replace="test"
$ echo ${var//12345678/$replace}
testabc
Not specific to the question, but for folks who need the same kind of functionality expanded for clarity from previous answers:
# create some variables
str="someFileName.foo"
find=".foo"
replace=".bar"
# notice the the str isn't prefixed with $
# this is just how this feature works :/
result=${str//$find/$replace}
echo $result
# result is: someFileName.bar
str="someFileName.sally"
find=".foo"
replace=".bar"
result=${str//$find/$replace}
echo $result
# result is: someFileName.sally because ".foo" was not found
Found a graceful solution.
echo ${LINE//12345678/$replace}
Single quotes are very strong. Once inside, there's nothing you can do to invoke variable substitution, until you leave. Use double quotes instead:
echo $LINE | sed -e "s/12345678/$replace/g"
Let me give you two examples.
Using sed:
#!/bin/bash
LINE="12345678HI"
replace="Hello"
echo $LINE | sed -e "s/12345678/$replace/g"
Without Using sed:
LINE="12345678HI"
str_to_replace="12345678"
replace_str="Hello"
result=${str//$str_to_replace/$replace_str}
echo $result
Hope you will find it helpful!
echo $LINE | sed -e 's/12345678/'$replace'/g'
you can still use single quotes, but you have to "open" them when you want the variable expanded at the right place. otherwise the string is taken "literally" (as #paxdiablo correctly stated, his answer is correct as well)
To let your shell expand the variable, you need to use double-quotes like
sed -i "s#12345678#$replace#g" file.txt
This will break if $replace contain special sed characters (#, \). But you can preprocess $replace to quote them:
replace_quoted=$(printf '%s' "$replace" | sed 's/[#\]/\\\0/g')
sed -i "s#12345678#$replace_quoted#g" file.txt
I had a similar requirement to this but my replace var contained an ampersand. Escaping the ampersand like this solved my problem:
replace="salt & pepper"
echo "pass the salt" | sed "s/salt/${replace/&/\&}/g"
use # if you want to replace things like /. $ etc.
result=$(echo $str | sed "s#$oldstr#$newstr#g")
the above code will replace all occurrences of the specified replacement term
if you want, remove the ending g which means that the only first occurrence will be replaced.
Use this instead
echo $LINE | sed -e 's/12345678/$replace/g'
this works for me just simply remove the quotes
I prefer to use double quotes , as single quptes are very powerful as we used them if dont able to change anything inside it or can invoke the variable substituion .
so use double quotes instaed.
echo $LINE | sed -e "s/12345678/$replace/g"
I'm trying to cut a string until it's first specified characters.
in this case, it's first Latin letter.
I tired this. It kind of works but sometimes shifts 1 or more characters.
f=$(echo $f | tail -c $((${#f}+-$(expr index "$f" [azertyuiopqsdfghjklmwxcvbnAZERTYUIOPQSDFGHJKLMWXCVBN])+2)))
take this string as example: ã%82¹ã%83%91ã%83¼ã%82¯ã%83«__original_ver.__-Your_name.mp3
I want to get: original_ver.__-Your_name.mp3
I tend to get this instead: ver.__-Your_name.mp3
Is there a better method? if so, some explanation is always welcome.
You can use extended globbing:
f=$(shopt -s extglob; LC_ALL=C; echo "${f##+([^[:alpha:]])}")
f=$(shopt -s extglob; LC_ALL=C; echo "${f/#+([^[:alpha:]])}")
or sed:
f=$(LC_ALL=C sed -r 's/^[^[:alpha:]]+//' <<< "$f")
Setting LC_ALL to C is mandatory, otherwise [[:alpha:]] might match wrong characters.
I want to add the string %%% to the beginning of some specific lines in a text file.
This is my script:
#!/bin/bash
a="c:\Temp"
sed "s/$a/%%%$a/g" <File.txt
And this is my File.txt content:
d:\Temp
c:\Temp
e:\Temp
But nothing changes when I execute it.
I think the 'sed' command is not finding the pattern, possibly due to the \ backslashes in the variable a.
I can find the c:\Temp line if I use grep with -F option (to not interpret strings):
cat File.txt | grep -F "$a"
But sed seems not to implement such '-F` option.
Not working neither:
sed 's/$a/%%%$a/g' <File.txt
sed 's/"$a"/%%%"$a"/g' <File.txt
I have found similar threads about replacing with sed, but they don't refer to variables.
How can I replace the desired lines by using a variable adding them the %%% char string?
EDIT: It would be fine that the $a variable could be entered via parameter when calling the script, so it will be assigned like:
a=$1
Try it like this:
#!/bin/sh
a='c:\\Temp' # single quotes
sed "s/$a/%%%$a/g" <File.txt # double quotes
Output:
Johns-MacBook-Pro:sed jcreasey$ sh x.sh
d:\Temp
e:\Temp
%%%c:\Temp
You need the double slash '\' to escape the '\'.
The single quotes won't expand the variables.
So you escape the slash in single quotes and pass it into the double quotes.
Of course you could also just do this:
#!/bin/sh
sed 's/\(.*Temp\)/%%%&/' <File.txt
If you want to get input from the command line you have to allow for the fact that \ is an escape character there too. So the user needs to type 'c:\\' or the interpreter will just wait for another character. Then once you get it, you will need to escape it again. (printf %q).
#!/bin/sh
b=`printf "%q" $1`
sed "s/\($b\)/%%% &/" < File.txt
The issue you are having has to do with substitution of your variable providing a regular expression looking for a literal c:Temp with the \ interpreted as an escape by the shell. There are a number of workarounds. Seeing the comments and having worked through the possibilities, the following will allow the unquoted entry of the search term:
#!/bin/bash
## validate that needed input is given on the command line
[ -n "$1" -a "$2" ] || {
printf "Error: insufficient input. Usage: %s <term> <file>\n" "${0//*\//}" >&2
exit 1
}
## validate that the filename given is readable
[ -r "$2" ] || {
printf "Error: file not readable '%s'\n" "$2" >&2
exit 1
}
a="$1" # assign a
filenm="$2" # assign filename
## test and fix the search term entered
[[ "$a" =~ '/' ]] || a="${a/:/:\\}" # test if \ removed by shell, if so replace
a="${a/\\/\\\\}" # add second \
sed -e "s/$a/%%%$a/g" "$filenm" # call sed with output to stdout
Usage:
$ bash sedwinpath.sh c:\Temp dat/winpath.txt
d:\Temp
%%%c:\Temp
e:\Temp
Note: This allows both single-quoted or unquoted entry of the dos path search term. To edit in place use sed -i. Additionally, the [[ operator and =~ operator are limited to bash.
I could have sworn the original question said replace, but to append, just as you suggest in the comments. I have updated the code with:
sed -e "s/$a/%%%$a/g" "$filenm"
Which provides the new output:
$ bash sedwinpath.sh c:\Temp dat/winpath.txt
d:\Temp
%%%c:\Temp
e:\Temp
Remember: If you want to edit the file in place use sed -i or sed -i.bak which will edit the actual file (and if -i.bak is given create a backup of the original in originalname.bak). Let me know if that is not what you intended and I'm happy to edit again.
Creating your script with a positional parameter of $1
#!/bin/bash
a="$1"
cat <file path>|sed "s/"$1"/%%%"$1"/g" > "temporary file"
Now whenever you want sed to find "c:\Temp" you need to use your script command line as follows
bash <my executing script> c:\\\\Temp
The first backslash will make bash interpret any backslashes that follows therefore what will be save in variable "a" in your executing script is "c:\\Temp". Now substituting this variable in sed will cause sed to interpret 1 backlash since the first backslash in this variable will cause sed to start interpreting the other backlash.
when you Open your temporary file you will see:
d:\Temp
%%%c:\Temp
e:\Temp
Is there a way to grep (or use another command) to find exact strings, using NO regex?
For example, if I want to search for (literally):
/some/file"that/has'lots\of"invalid"chars/and.triggers$(#2)[*~.old][3].html
I don't want to go through and escape every single "escapable". Essentially, I want to pass it through, like I would with echo:
$ echo "/some/file\"that/has'lots\of\"invalid\"chars/and.triggers$(#2)[*~.old][3].html"
/some/file"that/has'lots\of"invalid"chars/and.triggers$(#2)[*~.old][3].html
Use fgrep, it's the same as grep -F (matches a fixed string).
Well, you can put the information you want to match, each in a line, and then use grep:
grep -F -f patterns.txt file.txt
Notice the usage of the flag -F, which causes grep to consider each line of the file patterns.txt as a fixed-string to be searched in file.txt.
I want to check inside a file if it matches a binary pattern.
For that, I'm using clamAV signature database
Trojan.Bancos-166:1:*:3d415d736715ab5ee347238cacac61c7123fe35427224d25253c7b035558baf19e54e8d1a82742d6a7b37afc6d91015f751de1102d0a31e66ec33b74034b1ab471cc1381884dfdf0bb3e4233bd075fef235f342302ffd72ecabfa5aedf1b3dc99b3348346db4d9001026aef44c592fee61493f7262ad2bd1bce8a7ce60d81022533f6473ae184935f25cf6cc07c3aebfdf70a5a09139
I code this to retrieve the hex string representation signature
signature=$(echo "$line" |awk -F':' '{ print $4 }')
Moreover I change hex string to binary
printf -v variable $(sed 's/\(..\)/\\x\1/g;' <<< "$signature")
Until here It works perfectly.
Finally I would like to check if my file ( *$raw_file_path* ) matches my binary pattern (now in $variable)
I try this
test_var=$(grep -qU "$variable" "$raw_file_path")
or
test_var=$(grep -qU --regexp="$variable" "$raw_file_path")
I don't know why it doesn't work, Grep doesn't match anything
.
And sometimes some errors:
grep: Trailing backslash
grep: Invalid regular expression
I know it deals with pattern matching problems.
In my test I don't want use regular expression.
If you have any idea, or other bash tool.
Thanks.
You are currently using the --quiet option for grep by specifying q in -qU. This prevents grep from printing anything to stdout, therefore nothing will be saved to test_var.
Change your code to:
test_var=$(grep -UE "$variable" "$raw_file_path")
First the extra sub-shell can be avoided:
#!/bin/bash
signature="Trojan.Bancos-166:1:*:3d415d736715ab5ee347238cacac61c7123fe35427224d25253c7b035558baf19e54e8d1a82742d6a7b37afc6d91015f751de1102d0a31e66ec33b74034b1ab471cc1381884dfdf0bb3e4233bd075fef235f342302ffd72ecabfa5aedf1b3dc99b3348346db4d9001026aef44c592fee61493f7262ad2bd1bce8a7ce60d81022533f6473ae184935f25cf6cc07c3aebfdf70a5a09139"
variable=$(echo "${signature//*:/}" | sed 's/\(..\)/\\x\1/g;')
Require only confirmation of a match:
if grep -qU "$variable" "$raw_file_path"; then
# matches
fi
Or require the result for further processing:
test_var=$(grep -U "$variable" "$raw_file_path")
# contents of match in test_var
When returning to a variable, greps -q opt suppresses stdout
Edit
Tested working example
> signature="Trojan.Bancos-166:1:All_text before-the last : should be trimed:3d415d736715ab5ee347238cacac61c7123fe35427224d25253c7b035558baf19e54e8d1a82742d6a7b37afc6d91015f751de1102d0a31e66ec33b74034b1ab471cc1381884dfdf0bb3e4233bd075fef235f342302ffd72ecabfa5aedf1b3dc99b3348346db4d9001026aef44c592fee61493f7262ad2bd1bce8a7ce60d81022533f6473ae184935f25cf6cc07c3aebfdf70a5a09139" \
> hex_string=$( echo "${signature//*:/}" | sed 's/\(..\)/\\x\1/g;' ) \
> echo "$hex_string"
\x3d\x41\x5d\x73\x67\x15\xab\x5e\xe3\x47\x23\x8c\xac\xac\x61\xc7\x12\x3f\xe3\x54\x27\x22\x4d\x25\x25\x3c\x7b\x03\x55\x58\xba\xf1\x9e\x54\xe8\xd1\xa8\x27\x42\xd6\xa7\xb3\x7a\xfc\x6d\x91\x01\x5f\x75\x1d\xe1\x10\x2d\x0a\x31\xe6\x6e\xc3\x3b\x74\x03\x4b\x1a\xb4\x71\xcc\x13\x81\x88\x4d\xfd\xf0\xbb\x3e\x42\x33\xbd\x07\x5f\xef\x23\x5f\x34\x23\x02\xff\xd7\x2e\xca\xbf\xa5\xae\xdf\x1b\x3d\xc9\x9b\x33\x48\x34\x6d\xb4\xd9\x00\x10\x26\xae\xf4\x4c\x59\x2f\xee\x61\x49\x3f\x72\x62\xad\x2b\xd1\xbc\xe8\xa7\xce\x60\xd8\x10\x22\x53\x3f\x64\x73\xae\x18\x49\x35\xf2\x5c\xf6\xcc\x07\xc3\xae\xbf\xdf\x70\xa5\xa0\x91\x39